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The original Ackermann function $\varphi\colon \mathbb{N}\times\mathbb{N}\times\mathbb{N}_0\to \mathbb{N}$ as defined in [1] was invented to prove that there is a function that is recursive but not primitive recursive. It can be given by the following recursion:

  • $\varphi(a,b,0) = a+b$
  • $\varphi(a,b,n+1) = (x\mapsto \varphi(a,x,n))^b(\alpha(a,n))$

Where $\alpha(a,0)=0$, $\alpha(a,1)=1$ and $\alpha(a,n)=a$ for $n\ge 2$ are initial values and $(x\mapsto f(x))^k$ is the n-times k-times composition of the function $x\mapsto f(x)$. The function $n\mapsto \varphi(n,n,n)$ is not primitive recursive because - informally speaking - it grows too quickly.

These operations are right-bracedright-bracketed, this does not matter for $\varphi(a,b,1)=ab$ and $\varphi(a,b,2)=a^b$, but for the next higher rank it is important $\varphi(a,b,3)=\underbrace{a^\land (a^\land (...(a^\land a)))}_{b+1\; \text{occurences of}\; a}$ where $a^\land b:=a^b$.

If we would choose left-bracketing the functions would not grow so quickly. The left-bracketed operations would be defined as:

  • $\psi(a,b,0) = a+b$
  • $\psi(a,b,n+1) = (x\mapsto \psi(x,a,n))^b(\alpha(a,n))$

Again $\psi(a,b,1)=ab$ and $\psi(a,b,2)=a^b$, but here the forth operation would be $\psi(a,b,3)=a^{a^b}$

My question is now whether the left-bracketed operations still grow fast enough for not being primitive recursive, i.e. is $n\mapsto \psi(n,n,n)$ still not primitive recursive?

[1] Ackermann, W. (1928 ). Zum Hilbertschen Aufbau der reellen Zahlen. Math. Ann., 99, 118–133.
show/hide this revision's text 4 changed "braced" to "bracketed"

Left-braced Left-bracketed Ackermann function also not primitive recursive?

The original Ackermann function $\varphi\colon \mathbb{N}\times\mathbb{N}\times\mathbb{N}_0\to \mathbb{N}$ as defined in [1] was invented to prove that there is a function that is recursive but not primitive recursive. It can be given by the following recursion:

  • $\varphi(a,b,0) = a+b$
  • $\varphi(a,b,n+1) = (x\mapsto \varphi(a,x,n))^b(\alpha(a,n))$

Where $\alpha(a,0)=0$, $\alpha(a,1)=1$ and $\alpha(a,n)=a$ for $n\ge 2$ are initial values and $(x\mapsto f(x))^k$ is the n-times composition of the function $x\mapsto f(x)$. The function $n\mapsto \varphi(n,n,n)$ is not primitive recursive because - informally speaking - it grows too quickly.

These operations are right-braced, this does not matter for $\varphi(a,b,1)=ab$ and $\varphi(a,b,2)=a^b$, but for the next higher rank it is important $\varphi(a,b,3)=\underbrace{a^\land (a^\land (...(a^\land a)))}_{b+1\; \text{occurences of}\; a}$ where $a^\land b:=a^b$.

If we would choose left-bracing left-bracketing the functions would not grow so quickly. The left-braced left-bracketed operations would be defined as:

  • $\psi(a,b,0) = a+b$
  • $\psi(a,b,n+1) = (x\mapsto \psi(x,a,n))^b(\alpha(a,n))$

Again $\psi(a,b,1)=ab$ and $\psi(a,b,2)=a^b$, but here the forth operation would be $\psi(a,b,3)=a^{a^b}$

My question is now whether the left-braced left-bracketed operations still grow fast enough for not being primitive recursive, i.e. is $n\mapsto \psi(n,n,n)$ still not primitive recursive?

[1] Ackermann, W. (1928 ). Zum Hilbertschen Aufbau der reellen Zahlen. Math. Ann., 99, 118–133.
show/hide this revision's text 3 added 742 characters in body; edited tags; edited title; edited title

left braced hyper-operations Left-braced Ackermann function also not primitive recursive?

We know

The original Ackermann function $\varphi\colon \mathbb{N}\times\mathbb{N}\times\mathbb{N}_0\to \mathbb{N}$ as defined in [1] was invented to prove that for the there is a function that is recursive but not primitive recursive.It can be given by the following equations recursively defined operations $\uparrow_n\colon\mathbb{N}\times\mathbb{N}\to\mathbb{N}$:

  • $m \uparrow_1 k = m+k$recursion:

  • $m \uparrow_{n+1} 1 \varphi(a,b,0) = m$a+b$
  • $m \uparrow_{n+1} (k+1) \varphi(a,b,n+1) = m\uparrow_n (m x\mapsto \uparrow_{n+1} k)$varphi(a,x,n))^b(\alpha(a,n))$

    • Where $\alpha(a,0)=0$, $\alpha(a,1)=1$ and $\alpha(a,n)=a$ for $n\ge 2$ are initial values and $(x\mapsto f(x))^k$ is the n-times composition of the function $x\mapsto f(x)$.The function $n\mapsto n \uparrow_n n$ varphi(n,n,n)$ is not primitive recursive because - informally speaking - it grows too quickly.(the next 2

    These operations here are as expected $m\uparrow_2 k = mk$right-braced, this does not matter for $m\uparrow_3 k = m^k$)

    What do \varphi(a,b,1)=ab$ and $\varphi(a,b,2)=a^b$, but for the next higher rank it is important $\varphi(a,b,3)=\underbrace{a^\land (a^\land (...(a^\land a)))}_{b+1\; \text{occurences of}\; a}$ where $a^\land b:=a^b$.

    If we know about would choose left-bracing the functions would not grow so quickly. The left-braced operations?operations would be defined as:

  • $m \downarrow_1 k = m+k$
  • $m \downarrow_{n+1} 1 \psi(a,b,0) = m$a+b$
  • $m \downarrow_{n+1} (k+1) \psi(a,b,n+1) = (m\downarrow_{n+1} k) x\mapsto \downarrow_n m$

    • Here again $m\downarrow_2 k=mk$, psi(x,a,n))^b(\alpha(a,n))$

    Again $m\downarrow_3 k = m^k$ \psi(a,b,1)=ab$ and additionally $m\downarrow_4 k = m^{m^{k-1}}$.Is \psi(a,b,2)=a^b$, but here the function forth operation would be $\psi(a,b,3)=a^{a^b}$

    My question is now whether the left-braced operations still grow fast enough for not being primitive recursive, i.e. is $n\mapsto n\downarrow_n n$ again \psi(n,n,n)$ still not primitive recursive?

    [1] Ackermann, W. (1928 ). Zum Hilbertschen Aufbau der reellen Zahlen. Math. Ann., 99, 118–133.
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